Q11 of 97 Page 205

In ∆ABC, BD ⊥ AC and CE ⊥ AB such that BE=CD. Prove that BD=CE.

It is given that,

BD is perpendicular to AC and CE is perpendicular to AB

Now, in ∆BDC and ∆CEB we have:

BE = CD (Given)

∠BEC = ∠CDB = 90^o

And, BC = BC (Common)

∴ By RHS congruence rule

∆BDC ≅ ∆CEB

BD = CE (By CPCT)

Hence, proved

Prove that the perimeter of a triangle is greater than the sum of its three medians

Which is true?

(A) A triangle can have two acute angles.

(B) A triangle can have two right angles.

(D) An exterior angles of a triangle is always less than either of the interior opposite angles.

In ∆ABC, AB=AC. Side BA is produced to D such that AD=AB.

Prove that ∠BCD=90°.

In the given figure, it is given that AD=BC and AC=BD.

Prove that ∠CAD=∠CBD and ∠ADC=∠BCD.