EXERCISE 7.3 1. A ABC and A DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that (1) AABDA ACD (ii) ΔΑΒΡΞ Δ ACP (iii) AP bisects ZA as well as Z D. (iv) AP is the perpendicular bisector of BC. B A D P Fig. 7.39 C
Answers & Comments
Step-by-step explanation:
To prove the given statements, we will use the properties of isosceles triangles and corresponding angles.
(i) To prove ΔAABD ≅ ΔACD:
Since both triangles have A as a common vertex, we need to prove that AB = AC and ∠B = ∠C.
In triangle AABD and ACD,
AB = AD (isosceles triangle)
AC = AD (isosceles triangle)
∠A = ∠A (common)
So, by SAS congruence, ΔAABD ≅ ΔACD.
(ii) To prove ΔΑΒΡ ≅ ΔACP:
We need to show that AB = AC and ∠BAP = ∠CAP.
AB = AC (already proved in part (i))
To prove ∠BAP = ∠CAP:
∠BAP = ∠DAB (vertically opposite angles)
∠CAP = ∠DAC (vertically opposite angles)
In triangle AABD and ACD,
∠DAB = ∠DAC (isosceles triangle)
Therefore, ∠BAP = ∠CAP.
So, by SAS congruence, ΔΑΒΡ ≅ ΔACP.
(iii) To prove that AP bisects ∠A and ∠D:
In ΔAABD,
∠AAB + ∠ABD = 180° (sum of angles in a triangle)
∠A + ∠ABD = 180° (isosceles triangle, base angles are equal)
Similarly, in ΔACD,
∠ACD + ∠ADR = 180° (sum of angles in a triangle)
∠A + ∠ADR = 180° (isosceles triangle, base angles are equal)
From the above two equations, we can conclude that ∠ABD = ∠ADR.
Therefore, AP bisects ∠A and ∠D.
(iv) To prove that AP is the perpendicular bisector of BC:
From part (ii), we know that ΔΑΒΡ ≅ ΔACP.
Therefore, AB = AC and AP is the perpendicular bisector of BC.
Hence, AP is the perpendicular bisector of BC.