triangle ABC and triangle OBC are two isosceles triangle on the same base Bc and vertices plz answer it step by step I will make it branlist
Answers
Answer:
Step-by-step explanation:
In △ABC,AB=AC−−(1) [sides of an isosceles triangle are equal]
In △DBC,DB=DC−−(2) [sides of an isosceles triangle are equal]
In △ABD and △ACD
AB=AC [From (1)]
DB=DC [From (2)]
AD=AD [Common side]
△ABD≅△ACD
[by SSS Test of congruency ]
∴∠BAD=∠CAD(1)− [corresponding air of congurent triangle]
In △ABP and △ACP
AB=AC- [From (1)]
AP=AP- [Common side]
∠BAP=∠CAP- [From (3)]
△ABP≅△ACP- [by SAS test of congruency]
△BP=CP−−−(4) [corresponding pairs of congruency triangles]
∠APB=∠APL−−(5) [corresponding pairs of congruency triangles]
iii] In △BDP and △CDP
DB=DC- [From (1)]
BP=CP- [From (2)]
DP=DP- [Common side]
∴△BDP≅△CDP - [by SSS test of congruency]
∴∠BDP=∠CDP- [corresponding pair of congruent triangles]
Since ∠BDP=∠CDP we can say that
AP bisects ∠D
Since, ∠BAD=∠CAD- [From (3)]
we can say that AP bisects ∠A
iv] ∠APB+∠APC=180 degree- [Angle in a straight line is 180 ]
∴ ∠APB+∠APB=180 degree- [From (5)]
2∠APB=180 degree
∠APB=90 degree
∠APB=∠APC=90 degree
From, the above result, we can say that AP is the perpendicular bisector if BC