# Areas of Similar Triangles

In two similar triangles, the ratio of the areas of the triangles is equal to the ratio of the squares of the corresponding sides.**Given**

Î”ABC ~ Î”DEF

**To Prove**

**Construction**** **

Draw AG âŠ¥ BC and DH âŠ¥ EF.

**Proof**

=

Now in the Î”ABG and the Î”DEH

âˆ B = âˆ E

âˆ AGB = âˆ DHE (Each 90

^{o})

âˆ´Î”AGB ~Î”DHE

âˆ´

But,

âˆ´ (ii)

From the Equation (i) and (ii), we get

Similarly, we can prove

Hence,

The area of two similar triangles are in the ratio of the squares of the corresponding altitudes.

**Given **

Two similar triangles Î”ABC and Î”DEF where AL âŠ¥ BC and DM âŠ¥ EF. Also âˆ B = âˆ E, âˆ A = âˆ D

and âˆ C = âˆ F.

**To Prove**** **

**Proof**

Since

âˆ´ (i)

Now in Î”ABL and Î”DEM,

âˆ B = âˆ E (given)

âˆ ALB = âˆ DME (each 90

^{o})

âˆ´ Î”ABL ~ Î”DEM (AA similarity)

Hence,

â‡’ (ii)

From Equation (i) and (ii) we get,

**Example**

The areas of two similar triangles are in the ratio of the squares of the corresponding medians.

** Given **

Î”ABC ~ Î”DEF, and DL, AM are medians.

**To Prove **** **

**Proof**

Since Î”ABC ~ Î”DEF

Therefore, (i)

Also

â‡’ (ii)

and âˆ B = âˆ E

âˆ´Î”ABM ~ Î”DEL

Hence, (iii)

Thus, from Equations (ii) and (iii), we get

â‡’

(iv)Hence, from Equations (i) and (iv), we get

**Example **

P, Q are points on sides AB, AC respectively in Î”ABC and PQ Ã§Ã§ BC. Prove that the median AD bisects PQ.

Solution

**Given**

In the Î”ABC, AD is the median and PQ Ã§Ã§ BC.

**To Prove**

PR = QR

**Proof**

In Î”APR and Î”ABD

âˆ BAD = âˆ PAR (Common angles)

âˆ ARP = âˆ ADB (corresponding angle of || sides)

âˆ´ âˆ APR = âˆ ABD

âˆ´Î”APR ~ Î”ABD

âˆ´ (i) Similarly, (ii)

Comparing Equation (i) and (ii), we get,

But BD = CD [D is the mid-point of BC]

âˆ´ PR = QR

Thus the median AD bisects PQ.

**Example **

If two sides and a median bisecting the third side of a triangle are respectively proportional to the corresponding sides and the median of another triangle, then prove that the two triangles are similar.

**Solution**

**Given**

**Two triangles ABC and PQR in which AD and PS are median such that**

**To Prove**

Î”ABC ~ Î”PQR

**Construction**

Draw DE Ã§Ã§ AB and ST Ã§Ã§ PQ

**Proof**

In the Î”ABC, D is the mid-point of BC and DE Ã§Ã§ AB,

âˆ´DE= AB and E will be mid-point of AC.

Similarly, ST= PQ and T is mid-point of PR

We are given,

â‡’

or

â‡’ Î”ADE ~ Î”PST

âˆ´ âˆ DAE = âˆ SPT (i)

Similarly we can prove,

âˆ BAD = âˆ QPS (ii)

Adding Equations (i) and (ii) we get,

âˆ BAD + âˆ DAE = âˆ SPT + âˆ QPS

â‡’ âˆ BAD + âˆ DAC = âˆ SPR + âˆ QPS

â‡’ âˆ BAC = âˆ QPR

Also we are given

âˆ´Î”ABC ~ Î” PQR [SAS similarity of triangles]

**Example **

P is the mid-point of side BC of Î”ABC. If Q is the mid-point of AP and BQ when produced meets AC at L, prove that AL =

**Given **

DABC, in which P is the mid-point of BC and Q is the mid-point of AP

**To Prove**

AL =

**Construction**

Draw PO Ã§Ã§ BL meeting AC in O.

**Proof**

In the Î”BCL, PO Ã§Ã§ BL and P is the mid-point of BC.

.

^{.}. O is the mid-point of CL. Now in the Î”APO, QL Ã§Ã§ PO and Q is the mid-point of AP,

.

^{.}. L is the mid-point of AO

âˆ´AL = LO = OC

and AL + LO + OC = AC

â‡’ AL + AL + AL = AC

or AL=

**Example**

If the areas of two similar triangles are equal, then the two triangles will be congruent.

**Solution**

**Given**

Î”ABC ~ Î”DEF and ar (Î”ABC) = ar (Î”DEF)

**To Prove**

DABCâ‰…Î”DEF

**Proof**

âˆ´AB = DE, BC = EF and AC = DF

âˆ´Î”ABC â‰…Î”DEF [SSS Congruence]

**Example **

In the given figure, find x if AB Ã§Ã§ DC.

**Solution**

In the quadrilateral ABCD, AB || DC.

âˆ´ âˆ ABD = âˆ BDC and âˆ CAB = âˆ ACD

âˆ´ Î”ABO ~ Î”CDO

âˆ´

â‡’

or 3(3x - 19) = (x - 5)(x - 3)

or x^{2 }- 17x + 72 = 0

or (x - 9)(x - 8)=0

â‡’ x = 9 or 8

**Example **

In the fig. AB || CD || EF.Prove that

**Solution**

In Î”BCD, EF || CD

.....(1)

In Î”ABD, EF Ã§Ã§ AB

âˆ´

âˆ´

âˆ´ 1 -

âˆ´ 1 -

âˆ´ 1 =

âˆ´ 1 = EF

or

**Example **

In the fig. DEFG is a square and âˆ A = 90Â°

Prove that DE^{2 }= BD Â´ EC.

**Solution**

In Î”AGF and Î”DBG

âˆ GAF = âˆ BDG (each 90Â°)

âˆ AGF = âˆ GBD (Corresponding angles of Ã§Ã§ sides)

âˆ´ Î”AFG ~Î”DBG (By AA corollary)

Similarly, Î”AGF ~ Î”EFC

.

^{.}. Î”BDG ~ Î”FEC

Hence,

[DE = EF = FG = GD]

Hence, DE^{2 }= BD x EC

**Example**

In a || gm ABCD, the diagonal BD intersects the segment AE at F where E is the mid-point on BC, prove that DF = 2FB

In the || gm ABCD (fig). DC Ã§Ã§ AB and DA Ã§Ã§ CB.

**Solution**

In Î” ADF and Î”BEF

âˆ ADF = âˆ EBF (alternate angles)

âˆ DFA = âˆ EFB (Vertically Opposite angles)

.

^{.}. Î”A DF ~ Î”EBF

.

^{.}.

.

^{.}. â‡’

or

Hence, DF = 2 FB.