Q.1Triangle ABC is isoceles with AB=AC. D, E and F are midpoints of BC, AC and AB respectively. Show that line segments AD is perpendicular to the line segment EF and is bisected by it?

Q.2 If a diagonal of a parallelogram bisects one of the angles of the parallelogram, it also bisects the second angle and then the two diagonals are perpendicular to each other?

given: ABC is a triangle. and D, E and F are the mid-points of BC , AC and AB. and AB=AC

TPT: AD?EF and AG bisects EF.

proof:  construction: join EF.

in the triangle ADB and ADC,

BD=DC (D is the mid-point)


and AD is common.

therefore by SSS congruency triangle ADB is congruent to triangle ADC.

therefore ?BAD=?CAD therefore?FAG=?EAG

now in the triangle FAG and triangle EAG,


 AG is common.

and AF=AE [since E and F are the mid-points of AB and AC and AB=AC]

therefore by SAS congruency triangle FAG and triangle EAG are congruent.

hence by congruent property : FG=GE and ?AGF=?AGE=180/2=90 deg

hence AD bisects EF and perpendicular to EF.

hence proved

(2) given: PQRS is a parallelogram. and diagonal PR divides the ?P into two equal parts.

TPT: PR bisects the ?R and diagonals PR and QS are perpendicular.

proof: let PR and QS intersect at point O. 

let ?QPO=?OPS=x

since PS and QR are parallel lines. and PR is transversal line.


similarly for the set of parallel lines PQ and SR,

?QPR=?PRS=x, therefore ?PRQ=?PRS=x

hence PR bisects the ?R also.

in triangle PSR, ?SPR=?PRS=x

therefore PS=SR [sides opposite to equal angles are equal in length]

therefore PQRS is a rhombus.

and the diagonals of a rhombus are perpendicular to each other.

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