find a triangle 7.3
Answers
☯ SOLUTION:-
Given:- ∆ABC and ∆DBC are 2 isosceles triangles on the same base BC and vertices A and D are on the same side of BC. AD is extended to intersect BC at P.
To prove:-
i) ∆ ABD is congruent to ∆ACD
ii) ∆ABP is congruent to ∆ACP
iii) AP bisects angle A and as well as angle D.
iv) AP is the perpendicular bisector of BC.
Proof:-
i) In ∆ABD and ∆ACD,
AB= AC ....→(1) [ since ∆ABC is an iso. ∆le]
BD= CD ....→(2) [ since ∆DBC is an iso. ∆le]
AD = AD ....→(3) [Common ]
Therefore, by SSS rule ∆ABD is congruent to ∆ACD.
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ii) In ∆ABP and ∆ACP,
AB= AC ...(4) [From (1)]
angle ABP = angle ACP...(5) [Angles opposite to equal sides of a triangle are equal]
since ∆ABD is congruent to ∆ACD, [Proved in (i) above]
Therefore,
angle BAP = angle CAP ...(6) [C.P.C.T]
In view of (4), (5) and (6)
∆ABP is congruent to ∆ACP by ASA rule.
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iii) Since ∆ABP is congruent to ∆ACP
[Proved in (ii) above]
Therefore,
angle BAP = angle CAP [C.P.C.T]
=> AP bisects angle A.
In ∆BDP and ∆CDP
BD= CD ....(7) [From (2)]
DP= DP ...(8) [Common]
since, ∆ABP is congruent to ∆ACP [Proved in (ii) above]
Therefore,
BP = CP ...(9) [C.P.C.T]
In view of (7), (8), (9),
∆BDP is congruent to ∆CDP by SSS rule.
Therefore,
Angle BDP = angle CDP [C.P.C.T]
=> DP bisects angle D.
=> AP bisects angle D.
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iv) since ∆BDP is congruent to ∆CDP, [Proved in (iii) above]
Therefore,
BP = CP ....(10) [C.P.C.T]
angle BPD = angle CPD [C.P.C.T]
But, angle BPD + angle CPD = 180° [Linear pair axiom]
Therefore,
angle BPD = angle CPD = 90° ...(11)
In view of (10) and (11),
AP is the perpendicular bisector of BC.
HOPE IT HELPS UHH ♥ ✅
