practice problems

Q1. ABCD is a trapezium with side BC ∥ AD. If E is the midpoint of AB and the line through E parallel to DC meets AD and BC at X and Y respectively, prove that ABCD and XYCD have equal areas.

Q2. ABCD is a parallelogram and O is any point. The parallelograms OAEB, OBFC, OCGD, ODHA are completed. Show that EFGH is a parallelogram.

Q3. Let D be the midpoint of the side  BC of  triangle ABC. Prove that if AD > BD then angle A is acute; else if AD < BD, then angle A  is obtuse.

S3:
Since BD = CD, AD > BD => AD > CD
Let angle BAD = x and angle DAC = y and hence angle A = x + y.
In triangle ABD, AD > BD => Angle B > x
In triangle ACD, AD > CD => Angle C > y
B + C > x + y = A
B + C > A
A + B + C= 180 => A < 90 hence proved.
Similarly the other case can be proved.

S1: If I draw ABCD s.t. AD < BC then ABCD and XYCD have an overlapping pentagon between them which is EYCDA.
What is extra is this:
ABCD has triangle EBY
XYCD has triangle XEA.
If we show that they are congruent then we are done.
Angle BEY = Angle AEX
Angle XAE = Angle YBE
Hence all angles are same.
=> Triangles BEY ~ AEX
Sides AE = BE => they are congruent also.
H.P.

S2.
Below we show that O is inside ABCD but it could have been anywhere as the question said.
We will use vectors to solve this.
Let O be the origin.

In OAEB
vector OE = OA + OB => e = a + b

In OBFC
f = b + c
EF = FO + OE = e - f = a - c
Similarly we will find GH
In OCGD
OG = OC + OD = c + d = g
In ODHA
OH = h = OD + OA = a + d
GH = GO + OH = h - g = a - c

So EF and GH are parallel with same magnitude.
H.P.




S3

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