triangle ABC and DBC are two isosceles triangle on the same base BC and vertices A and D are on the same side of BC. if AD is extended to intersect BC and P, show that. (i) triangle ABD congruent to triangle (ii) triangle ABP is congruent to triangle ACP (iii) AP bisects angel A as well as angle D.( iv) AP is the perpendicular bisector of BC
Answers
Answer:
Step-by-step explaination
In Δ ABD and Δ ACD
AB =AC (ABC IS AN ISOSCELES TRIANGLE)
BD = DC (BDC IS AN ISOSCELES TRIANGLE)
AD = AD (COMMON)
Δ ABD ≅ ΔACD (SSS TEST )
∴ ∠ BAD = ∠CAD ( CPCT) (1)
∴ AD BISECTS ∠A (2)
IN Δ ABP AND Δ ACP
AB =AC ( ABC IS AN ISOSCELES Δ )
∠BAD = ∠CAD ( FROM (1) )
AD =AD (CPCT)
ΔABD ≅ ΔACD (SAS TEST )
BP = CP (CPCT) (3)
∠APB = ∠APC (CPCT) (4)
∠APB +∠APC =180° ( LINEAR PAIR)
∠APB +∠APB = 180° (FROM (4) )
2∠APB =180°
∠APB = 90° (5)
∴ AP IS THE PERPENDICULAR BISECTOR OF BC ( FROM (3) AND (5) )
IN Δ BDP AND Δ CDP
PD = PD (COMMON SIDE )
BP = CP (FROM (3) )
BD = CD ( BDC IS AN ISOSCELES Δ )
Δ BDP ≅ Δ CDP (SSS TEST)
∠ BDP = ∠ CDP (CPCT) (6)
AP BISECTS ∠A AS WELL AS ∠D ( FROM (2) AND (6) )
HENCE PROVED
To Show :
∆ABD ≅ ∆ACD.
∆ABP ≅ △ACP
AP bisects ∠A as well as ∠D
AP is perpendicular bisector of BC .
Proof :
In △ABD AND △ACD
AB = AC [ GIVEN]
AD = AD [ COMMON ]
BD = CD [ GIVEN ]
Therefore, ∆ABD ≅ ∆ACD.[SSS]
2. In △ABP and △ ACP
AB = AC [ GIVEN ]
∠BAP = ∠CAP [ C.P.C.T ]
AP = AP [ COMMON ]
Therefore , △ABP ≅ △ACP [ SAS ]
Other parts referred as attachment..
