Advertisements
Advertisements
प्रश्न
In the following figure, AB = EF, BC = DE and ∠B = ∠E = 90°.
Prove that AD = FC.
Advertisements
उत्तर
Given that, BC = DE
⇒ BC + CD = DE + CD ....( Adding CD on both sides )
⇒ BD = CE ....(i)
Now, in ΔABD and ΔFEC,
AB = EF ....(given)
∠ABD = ∠FEC ....(Each 90°)
BD = CE ....[ From (i) ]
⇒ ΔABD ≅ ΔFEC ...(by SAS congruence criterion)
⇒ AD = FC ...(c.p.c.t.)
APPEARS IN
संबंधित प्रश्न
In ΔABC, ∠A = 30°, ∠B = 40° and ∠C = 110°
In ΔPQR, ∠P = 30°, ∠Q = 40° and ∠R = 110°
A student says that ΔABC ≅ ΔPQR by AAA congruence criterion. Is he justified? Why or why not?
In the figure, the two triangles are congruent.
The corresponding parts are marked. We can write ΔRAT ≅ ?

In Δ ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in P. Arrange AP, BP and CP in descending order.
In the given figure, prove that:
CD + DA + AB + BC > 2AC

In two triangles ABC and DEF, it is given that ∠A = ∠D, ∠B = ∠E and ∠C =∠F. Are the two triangles necessarily congruent?
In the given figure, ABC is an isosceles triangle whose side AC is produced to E. Through C, CD is drawn parallel to BA. The value of x is

A triangle ABC has ∠B = ∠C.
Prove that: The perpendiculars from B and C to the opposite sides are equal.
In ∆ABC, AB = AC. Show that the altitude AD is median also.
ABC is an isosceles triangle with AB = AC and BD and CE are its two medians. Show that BD = CE.
ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.
