Triangles EX 6.3 Solutions
Question - 11 : - In the followingfigure, E is a point on side CB produced of an isosceles triangle ABC with AB =AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.
Answer - 11 : - 
Solution:
Given, ABC is an isosceles triangle.
∴ AB = AC
⇒ ∠ABD = ∠ECF
In ΔABD and ΔECF,
∠ADB = ∠EFC (Each 90°)
∠BAD = ∠CEF (Already proved)
∴ ΔABD ~ ΔECF (using AA similarity criterion)
Sides AB and BC andmedian AD of a triangle ABC are respectively proportional to sides PQ and QRand median PM of ΔPQR (see Fig 6.41). Show that ΔABC ~ ΔPQR.
Answer - 12 : - 
Solution
Question - 13 : - D is a point on the side BC of a triangle ABCsuch that ∠ADC= ∠BAC. Show that CA2 = CB.CD
Answer - 13 : -
Given, D is a point on the side BC of atriangle ABC such that ∠ADC = ∠BAC.
In ΔADC and ΔBAC,
∠ADC = ∠BAC (Already given)
∠ACD = ∠BCA (Common angles)
∴ ΔADC ~ ΔBAC (AA similarity criterion)
We know that corresponding sides of similar triangles are in proportion.
∴ CA/CB = CD/CA
⇒ CA2 = CB.CD.
Hence, proved.
Question - 14 : - Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.
Answer - 14 : -
Given: Two triangles ΔABC and ΔPQR in which ADand PM are medians such that;
AB/PQ = AC/PR = AD/PM
We have to prove, ΔABC ~ ΔPQR
Let us construct first: Produce AD to E sothat AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also JoinRN.
In ΔABD and ΔCDE, we have
AD = DE [By Construction.]
BD = DC [Since, AP is the median]
and, ∠ADB = ∠CDE [Vertically opposite angles]
∴ ΔABD ≅ ΔCDE [SAScriterion of congruence]
⇒ AB = CE [By CPCT] …………………………..(i)
Also, in ΔPQM and ΔMNR,
PM = MN [By Construction.]
QM = MR [Since, PM is the median]
and, ∠PMQ = ∠NMR [Vertically opposite angles]
∴ ΔPQM = ΔMNR [SAS criterion of congruence]
⇒ PQ = RN [CPCT] ………………………………(ii)
Now, AB/PQ = AC/PR = AD/PM
From equation (i) and (ii),
⇒CE/RN = AC/PR = AD/PM
⇒ CE/RN = AC/PR = 2AD/2PM
⇒ CE/RN = AC/PR = AE/PN [Since 2AD = AE and 2PM = PN]
∴ ΔACE ~ ΔPRN [SSS similarity criterion]
Therefore, ∠2 = ∠4
Similarly, ∠1 = ∠3
∴ ∠1 + ∠2 = ∠3 + ∠4
⇒ ∠A = ∠P …………………………………………….(iii)
Now, in ΔABC and ΔPQR, we have
AB/PQ = AC/PR (Already given)
From equation (iii),
∠A = ∠P
∴ ΔABC ~ ΔPQR [ SAS similarity criterion]
Question - 15 : - A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower
Answer - 15 : -
Given, Length of the vertical pole = 6m
Shadow of the pole = 4 m
Let Height of tower = h m
Length of shadow of the tower = 28 m
In ΔABC and ΔDEF,
∠C = ∠E (angular elevationof sum)
∠B = ∠F = 90°
∴ ΔABC ~ ΔDEF (AA similarity criterion)
∴ AB/DF = BC/EF (If two triangles are similar corresponding sidesare proportional)
∴ 6/h = 4/28
⇒h = (6×28)/4
⇒ h = 6 × 7
⇒ h = 42 m
Hence, the height of the tower is 42 m.
Question - 16 : - If AD and PM aremedians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQRprove that AB/PQ = AD/PM.
Answer - 16 : -
Given, ΔABC ~ ΔPQR
We know that the corresponding sides ofsimilar triangles are in proportion.
∴AB/PQ = AC/PR = BC/QR……………………………(i)
Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ………….…..(ii)
Since AD and PM are medians, they will divide their opposite sides.
∴ BD = BC/2 and QM = QR/2 ……………..………….(iii)
From equations (i) and (iii), we get
AB/PQ = BD/QM ……………………….(iv)
In ΔABD and ΔPQM,
From equation (ii), we have
∠B = ∠Q
From equation (iv), we have,
AB/PQ = BD/QM
∴ ΔABD ~ ΔPQM (SAS similarity criterion)
⇒AB/PQ = BD/QM = AD/PM