Practice problems

Q1. If the medians BE and CF are equal in a △ABC prove that AB=AC. 
S1.

Let G be the centroid and let CG = 2x = BG.

Let EG = FG = x.

Now triangles EGC and FGB are congruent by SAS. Why?

Since EG = FG, CG = BG and angle FGB = angle CGE because they are vertical angles.

=> FB = EC

And BF = FA since F is midpoint of AB, similarly AE = EC.

=> AF + FB = AE + EC => AB = AC H.P. Q2. Let (D, E, F) be the feet of the altitude from (A, B, C) in a (\triangle ABC). Prove that the perpendicular bisector of (EF) also bisects (BC).

S2. BCEF is a cyclic quadrilateral with BC as diameter. Perpendicular bisector of EF will pass through the center of the circle since EF is a chord. Center lies on the midpoint of BC as it is a diameter. Hence perpendicular bisector of EF will pass through the midpoint of BC. H.P. Q3. If $S$ is the circumcentre of a $\mathrm{△}ABC$ and $D,E,F$ are the feet of the altitudes of $\mathrm{△}ABC$ then prove that $SB\perp DF$. S3. Using properties of Orthic triangles.

We know that Angle FDB = A If S is the circumcenter then BSC = 2A and SB = SC => SCB = SBC = 90 - A In triangle DBX: Angle BXD = 180 - DBX - BDX = 180 - SBC - FDB = 90 H.P.
Q4. Let P be any point inside a regular polygon. If di is the distance of P from the ith side of the polygon, prove that d1 + d2 ... dn = constant, where n is the number of sides of the polygon. S4. Join P with each pair of adjacent vertices. Hence we get 'n' triangles area of which adds up to the polygon area. For e.g. in case of a triangle: [ABC] = a.d1/2 + a.d2/2 + a.d3/2 where a = regular polygon side. => Sigma(di) = [ABC].2/a We also know that [ABC] = s.r where s is semi-perimeter and r = inradius. => Sigma(di) = s.r.2/a Also s = 3a/2 => Sigma(di) = 3a/2.r.2/a = 3r In an equilateral triangle we can show that h(altitude) = 3r => Sigma(di) = h is known as Viviani's theorem. Similarly for regular n-gon it will be n.r Q5. In triangle $ABC$, let $P$ and $R$ be the feet of the perpendiculars from $A$ onto the external and internal bisectors of $\mathrm{\angle }ABC$, respectively; and let $Q$ and $S$ be the feet of the perpendiculars from $A$ onto the internal and external bisectors of $\mathrm{\angle }ACB$, respectively. If $PQ=7,QR=6$ and $RS=8$, what is the area of triangle $ABC$ ?
S5.

We know that feet of perpendiculars from A to internal/external angle bisectors of B,C are colinear. Now it's clear that APBR is a rectangle. Why? APB = 90, PBR = 90, BRA = 90 => the fourth angle will also be 90. In a rectangle, diagonals bisect each other and they are also equal => PR = AB = c => c = PR = PQ + QR = 13 Similarly, ASCQ is a rectangle and QS = AC = b. => b = QR + RS = 14 Now if we could only find BC = a we can use heron's formula to compute the area. Since PR bisects AB, let's say their intersection point is Mc. Similarly QS bisects AC at Mb. We know that McMb = BC/2 so we just need to find McMb. McMb = PS - PMc - MbS PS = PQ + QR + RS = 21 PMc = PR/2 = 6.5 MbS = QS/2 = 14/2 = 7 => McMb = 21 - 6.5 - 7 = 7.5 => BC = a = 15 Now we have all the sides so apply Heron's formula: Area = sqrt(s.(s-a)(s-b)(s-c)) s = 14 + 13 + 15)/2 = 21 sqrt(21.6.7.8) = 7.3.4 = 84 = Answer. Q6. For an acute triangle $\mathrm{△}ABC$ with orthocenter $H$, let $H_A$ be the foot of the altitude from $A$ to $BC$, and define $H_B$ and $H_C$ similarly. Show that $H$ is the incenter of $\mathrm{△}{H}_A{H}_B{H}_C$.

S6.

We know that Orthic triangle HaHbHc is anti parallel to ABC.
Angle HcHaHb = 180 - 2A
Angle HHaHc = HHaHb = 90 - A
So HHa is internal angle bisector of HcHaHb.
Similarly we can show for other angles of the Orthic triangle as well.

Q7.
Let $ABC$ be an acute angled scalene triangle with circumcentre $O$ and orthocentre $H$. If $M$ is the midpoint of $BC$, then show that $AO$ and $HM$ intersect on the circumcircle of $ABC$.

S7.
Earlier we proved that AH || OM and AH = 2OM.
Now let's assume AO and HM intersect at A'.
Since AH || OM, we can easily show what A'AH and A'OM are similar triangles.
AH/OM = 2/1 = AA'/OA'
Since AOA' is a straight line and we showed that O is the midpoint of AA' => OA = OA' = R = circumradius and AA' is a diameter of the circumcircle.
H.P.