In \(\triangle ABC\), an angle bisector from A meets BC at D. If AD bisects \(\angle BAC\), and AB = AC, are \(\triangle ABD\) and \(\triangle ACD\) congruent? If so, by what rule?
Yes, by SAS
Compare the two triangles ABD and ACD.
(i) \(AB = AC\) — given.
(ii) \(\angle BAD = \angle CAD\) — AD is the angle bisector.
(iii) \(AD = AD\) — common side.
So we have two sides and the included angle equal in the two triangles. This is the SAS criterion for congruence.
Hence \(\triangle ABD \cong \triangle ACD\) by SAS — option (2).
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Important Questions from Triangles, Congruence and Similarity
The radius of the circumcircle of an equilateral triangle of √3 unit side, is:
If the ratio of the angles of a triangle is 3 : 5 : 7, find the value of the largest angle.
A. 36°
B. 60°
C. 84°
D. 15°
If ΔABC and Δ PQR are similar and \(\rm\frac{BC}{QR} = \frac{1}{3}\) , find \(\rm\frac{ar(\Delta PQR)} {ar(\Delta BCA)}\)
If the angles of a triangle are in the ratio of 2 : 5 : 8, then find the value of the smallest angle.
ABCD is a parallelogram. Side BC is produced to E such that BC = CE. Join AE which intersects side CD at P. The area of triangle ABE is:
