 Hello and welcome to the session. I am Arsha and I am going to help you with the following problem that says triangle ABC and triangle DPC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that first is triangle ABT is congruent to triangle ACD, second is triangle ABP is congruent to triangle ACP, third is AP bisects angle A as well as angle D and fourth is AP is the perpendicular bisector of BC. Let us now begin with the solution. First let us write down what we are given. So here we are given two isosceles triangle ABC, BC and the common base BC such that side AB is equal to AC, this side is equal to this side and BD is equal to DC, this is equal to this side and this is because we are given that triangle ABC and BDC are isosceles triangle and in an isosceles triangle two sides are equal. Now we have to prove first a triangle ABD is congruent to triangle ADC then we have to prove triangle ABP is congruent to triangle ACP, third we have to prove that AP bisects angle A, angle D and lastly that AP is perpendicular bisector of BC. Now first we will try to show that triangle ABD is congruent to triangle ADC, in triangle ABD triangle ADC side AB is equal to side AC, these two sides this is given to us, triangle ABC is isosceles next side BT is equal to side TC, this is also given to us since triangle BDC is an isosceles triangle and AD is equal to AD since it is the common side. So this implies by side side side criteria triangle ABD is congruent to triangle ADC, this is the first part we have to prove. Now triangle ABD is congruent to triangle ADC this implies that angle BAD is equal to angle CAD, this is my CPCT which is the third part we have to prove that is angle BAD is equal to angle CAD, this is by CPCT. Now let us prove the second part and for that let us look in triangles ABP triangle ACP help AB is equal to AC, this is given to us also AP is equal to AP since this is the common side angle BAD is equal to angle CAD this just now we have proved by third part from here. Now there is C AB is equal to AC, AP is equal to AP and this is the included angle therefore this implies that triangle ABP is congruent to triangle ACP and this is by SAS criteria this further implies that angle ABP is equal to angle APC this is by CPCT and ABP is this angle and APC is this angle these two angles are equal also BP is equal to Cp this is again by CPCT so this is what we have to prove in the second part and now let us show that AP is the perpendicular bisector of BC now sum of angles APB and angle APC is equal to 180 degree since they form a linear pair this implies two times angle APB is equal to 180 degree which further implies that angle APB is equal to 180 degree upon 2 which is further equal to 90 degree so this implies angle APB is equal to 90 degree which is this angle and also BP is equal to PC this is again since we have proved above and so this implies that AP is perpendicular bisector of line BC in the third part we have to show that AP bisects angle A and angle D we have shown that AP bisects angle A this is because angle BAD is equal to angle CAD since the triangles ABD and triangle ADCR congruent and now we have to show that AP bisects angle D so what we will do is we will try to show these two triangles congruent so in triangle B DP and triangle CDP we find BP is equal to PC we have proved them third part also DPB is equal to angle DPC which is equal to 90 degree and BT is equal to DC this is given to us and these two we have proved above and this implies that by RHS criteria triangle B DP is congruent to triangle CDP which further implies that angle B DP is equal to angle CDP and this is by CPCT which further implies that AP bisects angle D this completes the proof of all the fall parts hope you enjoyed it take care and have a good day