Question

in the adjoining figure AL and CM are perpendicular to the diagonal BD of a parallelogram ABCD prove that:

(i) triangle ALD is conruent to triangle CMB

(ii)AL=CM​

Answer 1

PROOVED!

POINTS TO REMEMBER -

•OPPOSITE SIDES OF PARALLELOGRAM ARE EQUAL

•OPPOSITE ANGLES OF PARALLELOGRAM ARE EQUAL

• ALTERNATE ANGLES ARE EQUAL.

• WHEN TWO ANGLES AND 1 SIDES ARE EQUAL THEN WE HAVE TO APPLU AAS CONGURENCY

Answer 2

Given,

• AL and CM are perpendiculars to the diagonal BD.
• ABCD is a parallelogram.

To find,

We have to prove that

• ΔALD≅ΔCMB

• AL= CM

Solution,

In the adjoining figure, AL and CM are perpendicular to the diagonal BD of a parallelogram ABCD, then ΔALD≅ΔCMB by AAS congruency rule and AL=CM by CPCT.

   In ΔALD and ΔCMB

             AD = BC (opposite sides of parallelogram are equal)   (1)

             ∠ B = ∠ D ( opposite angles of parallelogram)

        1/2∠ B = 1/2∠ D

        ∠CBM = ∠ADL        (2)

         ∠ALD = ∠CMB (each 90°)  (3)

From (1),(2), and (3),

     ΔALD ≅ ΔCMB (By AAS congruency rule)

Now,    In ΔALD and ΔCMB

              ΔALD ≅ ΔCMB

        Then,  AL = CM (By CPCT)

Hence, in the adjoining figure, AL and CM are perpendicular to the diagonal BD of a parallelogram ABCD, then ΔALD≅ΔCMB by AAS congruency rule and AL=CM by CPCT.