In triangle abc the base ab …

Given:- In {tex}\Delta{/tex}ABC, CA = CB and AP {tex}\times{/tex} BQ = AC²

To prove:- {tex}\Delta{/tex}APC ~ {tex}\Delta{/tex}BCQ
Proof:- AP {tex}\times{/tex} BQ = AC² [Given]
{tex}\Rightarrow AP \times BQ = AC \times AC{/tex}
{tex} \Rightarrow AP \times BQ = AC \times BC{/tex} [AC = BC given]
{tex}\Rightarrow \frac{{AP}}{{BC}} = \frac{{AC}}{{BQ}}{/tex} ....(i)
Since, CA = CB [Given]
Therefore, {tex}\angle CAB = \angle CBA{/tex} ...(ii) [Opposite angles to equal sides]
Now, {tex}\angle CAB + \angle CAP = 180^\circ {/tex} ...(iii) [Linear pair of angles]
And, {tex}\angle CBA + \angle CBQ = 180^\circ {/tex} ...(iv) [Linear pair of angles]
Therefore, Comparing  equation (ii)(iii) and (iv),we get,
{tex}\angle CAP = \angle CBQ{/tex} ...(v)
In {tex}\Delta{/tex}APC and {tex}\Delta{/tex}BCQ
{tex}\angle CAP = \angle CBQ{/tex} [From (v)]
{tex}\frac{{AP}}{{BC}} = \frac{{AC}}{{BQ}}{/tex} [From (i)]

Therefore,by SAS criteria of similar triangles,we have,
Then, {tex}\Delta{/tex}APC ~ {tex}\Delta{/tex}BCQ