Advertisements
Advertisements
Question
In quadrilateral ABCD, AD = BC and BD = CA.
Prove that:
(i) ∠ADB = ∠BCA
(ii) ∠DAB = ∠CBA
Solution
Given: In quadrilateral ABCD, AD = BC and BD = AC.
To Prove:
(i) ∠ADB = ∠BCA
(ii) ∠DAB = ∠CBA
Proof:
In ΔABD and ΔBAC,
AD = BC ....(given)
BD = CA ....(given)
AB = AB ....(common)
∴ ΔABD ≅ ΔBAC ....(by SSS congruence criterion)
`{:(∠"ADB" = ∠"BCA"), (∠"DAB" = ∠"CBA"):}} ...("c.p.c.t.")`
APPEARS IN
RELATED QUESTIONS
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that:
- ΔAMC ≅ ΔBMD
- ∠DBC is a right angle.
- ΔDBC ≅ ΔACB
- CM = `1/2` AB
You have to show that ΔAMP ≅ AMQ.
In the following proof, supply the missing reasons.
| Steps | Reasons | ||
| 1 | PM = QM | 1 | ... |
| 2 | ∠PMA = ∠QMA | 2 | ... |
| 3 | AM = AM | 3 | ... |
| 4 | ΔAMP ≅ ΔAMQ | 4 | ... |
ABCD is a square, X and Yare points on sides AD and BC respectively such that AY = BX. Prove that BY = AX and ∠BAY = ∠ABX.
ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.
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
D, E, F are the mid-point of the sides BC, CA and AB respectively of ΔABC. Then ΔDEF is congruent to triangle
A triangle ABC has ∠B = ∠C.
Prove that: The perpendiculars from the mid-point of BC to AB and AC are equal.
From the given diagram, in which ABCD is a parallelogram, ABL is a line segment and E is mid-point of BC.
Prove that: AB = BL.
In the adjoining figure, QX and RX are the bisectors of the angles Q and R respectively of the triangle PQR.
If XS ⊥ QR and XT ⊥ PQ ;
prove that: (i) ΔXTQ ≅ ΔXSQ.
(ii) PX bisects angle P.
In ΔABC, AB = AC and the bisectors of angles B and C intersect at point O.
Prove that : (i) BO = CO
(ii) AO bisects angle BAC.
