Question

So,
Δ PAB = Δ PAC
(RHS rule)
(CPCT)
So,
Z PAB = Z PAC
ou have been mainly
s. Sometimes, we do
mple, line-segment A
1 (i) and Z A is great
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) A ABD=AACD
(1) A ABP=AACP
(IN AP bisects Z A as well as Z D.
B
(iv) AP is the perpendicular bisector of BC.
Fig. 13
t us now examine
al angles of a trian
Py: Fix two pins
C of a triangle
x one end of anot
other (free) en
and draw A A
1 AD bisects BC
(ii) AD bisects ZA.
3. Two sides AB and
Note that this result is the converse of the result proved in Q.501
2. AD is an altitude of an isosceles triangle ABC in which AB = ACS​

Answer 1

Answer:

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Step-by-step explanation:

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