In an isosceles ΔABC, the base AB is produced both the ways to P and

1.	In an isosceles ΔABC, the base AB is produced both the ways to P and Q such that AP × BQ = AC2. Prove that .ΔAPC ~ BCQ,
Answer» Given : In ΔABC , CA – CB and AP x BQ = AC² To prove :- ΔAPC ~ BCQ Proof:- AP X BQ = AC² (Given) Or, AP x BC = AC x AC Or, AP x BC = AC x BC (AC = BC given) Or, AP/BC = AC/PQ ………………(i) Since, CA = CB (Given) Then,∠CAB = ∠CBA …………….(ii) (Opposite angle to equal sides) NOW ∠CAB + ∠CAP = 180° …………(iii) (Linear pair of angle) And ∠CBA + ∠CBQ = 180° …………..(iv) (Linear pair of angle) Compare equation (ii) (iii) & (iv) ∠CAP = ∠CBQ ……………..(v) In ΔAPC and ΔBCQ ∠CAP = ∠CBQ (From equation v) AP/BC = AC/PQ (From equation i) Then , ΔAPC ~ ΔBCQ (By SAS similarity)

In an isosceles ΔABC, the base AB is produced both the ways to P and Q such that AP × BQ = AC2. Prove that .ΔAPC ~ BCQ,

Answer»

Given : In ΔABC , CA – CB and AP x BQ = AC²

To prove :- ΔAPC ~ BCQ

Proof:-

AP X BQ = AC² (Given)

Or, AP x BC = AC x AC

Or, AP x BC = AC x BC (AC = BC given)

Or, AP/BC = AC/PQ ………………(i)

Since, CA = CB (Given)

Then,∠CAB = ∠CBA …………….(ii) (Opposite angle to equal sides)

NOW ∠CAB + ∠CAP = 180° …………(iii) (Linear pair of angle)

And ∠CBA + ∠CBQ = 180° …………..(iv) (Linear pair of angle)

Compare equation (ii) (iii) & (iv)

∠CAP = ∠CBQ ……………..(v)

In ΔAPC and ΔBCQ

∠CAP = ∠CBQ (From equation v)

AP/BC = AC/PQ (From equation i)

Then , ΔAPC ~ ΔBCQ (By SAS similarity)