Advertisements
Advertisements
Question
ΔABC is isosceles with AB = AC. If BC is extended to D, then prove that AD > AB.
Advertisements
Solution
Using the exterior angle and property in ΔACD, we have
∠ACB = ∠CDA + ∠CAD
⇒ ∠ACB > ∠CDA ------(1)
Now, AB = AC
Now, ∠ACB = ∠ABC ------(2)
From (1) and (2)
∠ABC > ∠CDA
It is known that, in a triangle, the greater angle has the longer side opposite to it.
Now, In ΔABD, er have ∠ABC > ∠CDA
∴ AD > AB.
APPEARS IN
RELATED QUESTIONS
ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.
Complete the hexagonal and star shaped rangolies (see the given figures) by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?
From the following figure, prove that: AB > CD.
In a triangle PQR; QR = PR and ∠P = 36o. Which is the largest side of the triangle?
In the following figure, write BC, AC, and CD in ascending order of their lengths.
"Issues of caste discrimination began to be written about in many printed tracts and essays in India in the late nineteenth century." Support the statement with two suitable examples.
Name the smallest angle in each of these triangles:
In ΔXYZ, XY = 6.2cm, XY = 6.8cm and YZ = 5cm
In ABC, P, Q and R are points on AB, BC and AC respectively. Prove that AB + BC + AC > PQ + QR + PR.
ABCD is a trapezium. Prove that:
CD + DA + AB > BC.
In ΔPQR, PS ⊥ QR ; prove that: PQ + PR > QR and PQ + QR >2PS.
