Question

ABCD is a trapezium in which ab parallel to dc and its diagonals intersect each other at point O if ab parallel to CD, Ho A is equals to X + 5 Y is equals to x minus 1 is equal to X + 3 and OD is equals to x minus 2 then find the value of x ​

Answer 1

(i) ∠DAB=∠CBA

Construction :- to produce AB to E and draw CE∥DA,MO∥AB

Proof: So AECD is a parallelogram.

In ΔBEC

BC=CE (AD=BC=CE,opposite sides of parallelogram)

∠CBE=∠CEB (opposite to equal sides) ---- (i)

∠ADC=∠BEC (opposite angle of a parallelogram) ----- (ii)

∠DAB+∠ADC=180

∘

(sum of co-interior angles) ------ (iii)

∠ABC=∠EBC=180

∘

(L.P.P.)

∠ABC+∠ADC=180

∘

---- (iv) from (ii)

From (iii) and (iv)

∠DAB=∠CBA

(ii) ∠BCD=∠EBC ---- (v) (alternate interior angles)

From (i),(ii), and (v)

∠ADC=∠BCD proved

(iii) In ΔABD and ΔBAC

∠BAD=∠ABC

AB=AB (common)

AD=BC (Given)

ΔABD≅ΔBAC (by SAS rule)

BD=AC (By C.P.C.T)

(iv) In ΔABD

MO∥AB

MD

AM

=

OD

BO

--- A, (by lemma of B.P.T)

In ΔADC

MO∥DC

MD

AM

=

OC

AO

---- B (by B.P.T.)

From A and B

OD

BO

=

OC

AO

---- C

OD

BO

+1=

OC

AO

+1

OD

BO+DO

=

OC

AO+OC

OD

BD

=

OC

AC

OD=OC (AC=BD)

From C

OB=OA (OD=OC)

Step-by-step explanation:

hope it will help you.....