Question

In an isosceles \[\Delta ABC\] , the use \[AB\] is produced both the ways to \[P\] and \[Q\] such that \[AP \times BQ = {\left( {AC} \right)^2}\] . Prove that \[\Delta APC \sim \Delta BCQ\] .

Answer 1

Hint: We have to prove that both the \[\Delta APC\] and \[\Delta BCQ\] are two similar triangles . We solve this question using the concept of similar triangles . We should have the knowledge of the definition of similar triangles, its properties and the various properties of angles of a triangle . First using the given relation we will simplify it such that we get the relation as ratios of the sides of the two triangles . And then using the properties of angle of the triangle with equal sides we will simplify the values of the angles to obtain a relation and then using the two relation one of the sides and the other of the angles we will prove that the two triangles are similar .

Complete step-by-step solution:
Given :
\[AP \times BQ = {\left( {AC} \right)^2}\]
\[AC = BC\]
To prove : \[\Delta APC \sim \Delta BCQ\]
Proof :
We can simplify the given relations as :
\[\dfrac{{AP}}{{BC}} = \dfrac{{AC}}{{BQ}} - - - - (1)\]
Since , we are given that :
\[AC = BC\]
We also know that angles opposite equal sides are also equal .
Using this property for angle , we get
\[\angle CAB{\text{ }} = \angle CBA---\left( 2 \right)\] [opposite angle of equal sides are equal]
Now , we also know that the sum of two angles on a straight line is always \[{180^\circ }\] .
So , we can also write the expression for the angles on the straight line as :
\[\angle CAB + \angle CAP = {180^\circ }---\left( 3 \right)\]
\[\angle CBA + \angle CBQ = {180^\circ } - - - - \left( 4 \right)\]
Now , on subtracting equation \[\left( 3 \right)\] from equation \[\left( 4 \right)\] , we get
\[\angle CAB + \angle CAP - \left( {\angle CBA + \angle CBQ} \right) = {180^\circ } - {180^\circ }\]
As we know that ∠CAB = ∠CBA , we can write the expression as :
\[\angle CAP - \angle CBQ = 0\]
On simplifying , we get
\[\angle CAP = \angle CBQ - - - - (5)\]
Now , in \[\Delta APC\] and \[\Delta BCQ\]
\[\angle CAP = \angle CBQ\] [ proved above ]
\[\dfrac{{AP}}{{BC}} = \dfrac{{AC}}{{BQ}}\] [from \[\left( 1 \right)\]]
\[\Delta APC \sim \Delta BCQ\]
[by Side Angle Side (SAS) similarity]
Hence proved that \[\Delta APC \sim \Delta BCQ\] .

Note: If there are two or more angles on a straight line without any common part and without leaving a space between the angles then , the sum of the angles is always equal to \[{180^\circ }\] . [the angle should be continuous as given in the question for the pair of angles \[\angle CAB\] and \[\angle CAP\] , angles \[\angle CBP\] and \[\angle CBQ\].
If a pair of angles of two triangles are equal in measure and the ratios of their corresponding sides are equal then the two triangles are said to be similar triangles . Which we have concluded in the above statements .