In an isosceles $\triangle ABC$, the base AB is produced both the ways to P and Q such that $AP \times BQ = AC^2$. Prove that $\triangle APC \sim \triangle BCQ$.
Given:
In an isosceles $\triangle ABC$, the base AB is produced both the ways to P and Q such that $AP \times BQ = AC^2$.
To do:
We have to prove that $\triangle APC \sim \triangle BCQ$.
Solution:
$AC=BC$ and $AP \times BQ = AC^2$.
$AP \times BQ = AC \times AC$
$AP \times BQ = AC \times BC$
$\frac{AP}{AC}=\frac{BC}{BQ}$.....(i)
In $\triangle ABC$
$\angle CAB=\angle CBA$....(ii) (Angles opposite to equal sides are equal)
$\angle CAB+\angle CAP=180^o$....(iii) (Linear pair)
$\angle CBA+\angle CBQ=180^o$....(iv) (Linear pair)
From (ii), (iii) and (iv), we have,
$\angle CAP=\angle CBQ$....(v)
In $\triangle APC$ and $\triangle BCQ$, from (i) and (v),
$\angle CAP=\angle CBQ$
$\frac{AP}{AC}=\frac{BC}{BQ}$
This implies,
$\triangle APC \sim \triangle BCQ$. (By SAS similarity)
Hence proved.
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