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)`
Answer: 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
Answer: 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
Answer: 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.
Answer: In Δ ABD and Δ ECF
∠ADB = ∠EFC (Right angle)
∠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
Answer: In triangle Δ ABC and Δ PQR
`(AB)/(PQ)=(AC)/(PR)=(AD)/(PM)` (given) ---------(1)
Hence; ∠BAD = ∠QPM
∠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.
Answer: In ΔBAC and ΔADC;
∠BAC = ∠ADC (given)
∠ACB = ∠DCA (Common angle)
Hence; ΔBAC ∼ ΔADC
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.
Answer: 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)`
Answer: 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