In ∆BEC and ∆CFB,
∠BEC = ∠CFB (Each 90°)
BC = CB (Common)
BE = CF (Given)
∠∆BEC ≅ ∠∆CFB (By RHS congruency)
∠BCE = ∠CBF (By CPCT)
∠AB = AC (Sides opposite to equal angles of a triangle are equal)
Hence, ∆ABC is isosceles.
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 (see Fig. 7.40). Show that:
(i) ∆ BM≅∆ PQN
(ii) ∆ ABC≅∆ PQR
In Fig. 9.26, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠ BAC.