(i) ∆ABD ≅ ∆ACD
(ii) ∆ABP ≅ ∆ACP
(iii) AP bisects ∠A as well as ∠D
(iv) AP is the perpendicular bisector of BC.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠A.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of triangle PQR (see figure). Show that:
(i) ∆ABM ≅ ∆PQN
(ii) ∆ABC ≅ ∆PQR.
Given: ABC is an isosceles triangle with AB = AC.
To Prove: ∠B = ∠C
Construction: Draw AP π BC
Proof: In right triangle APB and right triangle
APC,
Hyp. AB = Hyp. AC | Given
Side AP = Side AP | Common
∴ ∆APB ≅ ∆APC | RHS Rule
∴ ∠ABP = ∠ACP | C.P.C.T.
⇒ ∠B = ∠C.
The diagonals PR and QS of a quadrilateral PQRS intersect each other at O. Prove that
(i) PQ + QR + RS + SP > PR + QS
(ii) PQ + QR + RS + SP < 2 (PR + QS)