# Triangle

## NCERT Exercise 6.3

### Part 3

Question 10: CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of Δ ABC and Δ EFG respectively. If Δ ABC ∼ Δ FEG, show that:

(a) (CD)/(GH)=(AC)/(FG)

Solution: Δ ABC ∼ Δ FEG (given)

Hence; ∠ABC = ∠FEG

∠ACB = ∠FGE

∠BAC = ∠GFE -------------(1)

Hence; ∠ACD = ∠FGH (Halves of ∠BCA and ∠FGE) -------------(2)

From equations (1) and (2);

ΔACD ∼ Δ FGH

Hence; in these triangles,

(CD)/(AC)=(GH)/(FG)

Or, (CD)/(GH)=(AC)/(FG)

Proved

(b) Δ DCB ∼ Δ HGE

Solution: In Δ DCB and Δ HGE

∠DCB = ∠HGE (Halves of ∠BCA and ∠FGE)

∠DBC = ∠HEG (Because they are common to ∠ABC and ∠FEG)

Hence; Δ DCB ∼ Δ HE proved

(c) Δ DCA ∼ Δ HGF

Solution: In Δ DCA and Δ HGF

ΔACD ∼ Δ FGH [proved in question (a)]

Hence; ΔACD ∼ Δ FGH proved

(Note: points C and G are in middle and sequence of other points has been changed in naming these triangles. Hence; triangles in question (a) are similar in this question.)

Question 11: In the given figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that Δ ABD ∼ Δ ECF.

Solution: In Δ ABD and Δ ECF

∠ABC = ∠ECF (Angles opposite to equal sides)

Hence; Δ ABD ∼ Δ ECF (AAA criterion)

Question 12: Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of Δ PQR. Show that Δ ABC ∼ Δ PQR

Solution: In triangle Δ ABC and Δ PQR

(AB)/(PQ)=(AC)/(PR)=(AD)/(PM) (given) ---------(1)

∠DAC = ∠MPR

(These are angles made by a side and median of one triangle and corresponding side and median of another triangle)

Hence; ∠BAD + ∠DAC = ∠QPM + ∠MPR

Or, ∠BAC = ∠QPR -----------(2)

From equation (1) and (2);

Δ ABC ∼ Δ PQR proved (SAS criterion)

Question 13: D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD.

∠ACB = ∠DCA (Common angle)

Hence; (CA)/(CB)=(CD)/(CA)

(corresponding sides are in same ratio)

Or, CA xx CA = CB xx CD

Or, CA^2 = CB xx CD proved

Question 14: A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Solution: Height of pole = AB = 6 m and its shadow = BC = 4 m

Height of tower = PQ = ? and its shadow = QR = 28 m

The angle of elevation of the sun will be same at a given time for both the triangles.

Hence; ΔABC ∼ ΔPQR

This means;

(AB)/(AC)=(PQ)/(QR)

Or, 6/4=(PQ)/(28)

Or, PQ=(6xx28)/(4)=42 cm

Height of tower = 42 m

Question 15: If AD and PM are medians of triangles ABC and PQR, respectively where Δ ABC ∼ Δ PQR, prove that (AB)/(PQ)=(AD)/(PM)

Solution: Δ ABC ∼ Δ PQR (Given)

Hence; (AB)/(AD)=(PQ)/(PM)

(A side and the median of one triangle are in same ratio as a corresponding side and median of another triangle)

(AB)/(PQ)=(AD)/(PM)

Proved