Short Answer Type
Long Answer Type
(i) ∆ABD ≅ ∆ACD
(ii) ∆ABP ≅ ∆ACP
(iii) AP bisects ∠A as well as ∠D
(iv) AP is the perpendicular bisector of BC.
Short Answer TypeAD 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: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆PQR.
To Prove: (i) ∆ABM ≅ ∆PQN
(ii) ∆ABC ≅ ∆PQR.
Proof: (i) In ∆ABM and ∆PQN,
AB = PQ ...(1) | Given
AM = PN ...(2) | Given
BC = QR
⇒ 2BM = 2QN
| ∵ M and N are the mid-points of BC and QR respectively
⇒ BM = QN ...(3)
In view of (1), (2) and (3),
∆ABM ≅ ∆PQN | SSS Rule
(ii) ∵ ∆ABM ≅ ∆PQN
| Proved in (1) above
∴ ∠ABM = ∠PQN | C.P.C.T.
⇒ ∠ABC = ∠PQR ...(4)
In ∆ABC and ∆PQR,
AB = PQ | Given
BC = QR | Given
∠ABC = ∠PQR | From (4)
∴ ∆ABC ≅ ∆PQR. | SAS Rule
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)
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