Advertisements
Advertisements
प्रश्न
In a triangle, ABC, AB = BC, AD is perpendicular to side BC and CE is perpendicular to side AB.
Prove that: AD = CE.
Advertisements
उत्तर
ln ΔABD and ΔCBE,
AB = BC ....(given)
AD ⊥ BC
CE ⊥ AB
To proved:
In ΔABD & ΔCBE
∠ ADB = ∠ CEB = 90° ....[Perpendiculars]
∠B = ∠B ....(Common angle)
AB = BC
∴ ΔABD ≅ ΔCBE ....(by AAS congruence)
⇒ AD = CE ...(c.p.c.t.c)
APPEARS IN
संबंधित प्रश्न
You want to show that ΔART ≅ ΔPEN,
If you have to use SSS criterion, then you need to show
1) AR =
2) RT =
3) AT =
In the figure, the two triangles are congruent.
The corresponding parts are marked. We can write ΔRAT ≅ ?
Prove that the perimeter of a triangle is greater than the sum of its altitudes.
In the given figure, prove that:
CD + DA + AB + BC > 2AC
In two congruent triangles ABC and DEF, if AB = DE and BC = EF. Name the pairs of equal angles.
A triangle ABC has ∠B = ∠C.
Prove that: The perpendiculars from B and C to the opposite sides are equal.
In quadrilateral ABCD, AD = BC and BD = CA.
Prove that:
(i) ∠ADB = ∠BCA
(ii) ∠DAB = ∠CBA
A point O is taken inside a rhombus ABCD such that its distance from the vertices B and D are equal. Show that AOC is a straight line.
In the following figure, ∠A = ∠C and AB = BC.
Prove that ΔABD ≅ ΔCBE. 
PQRS is a parallelogram. L and M are points on PQ and SR respectively such that PL = MR.
Show that LM and QS bisect each other.
