# prove that altitude of a triangle r concurrent no

*print*Print*list*Cite

### 2 Answers

Prove that the altitudes of a triangle are congruent.

On a coordinate grid, let the three vertices be given by (-a,0), (a,0), and (b,c).

The side opposite (-a,0) has slope `(c-0)/(b-a)` , and since the altitude to that side will be perpendicular it will have the opposite reciprocal slope so `m=(a-b)/c` . Using this slope with the point (-a,0) gives the equation for this altitude as `y-0=(a-b)/c(x-(-a))` or `y=(a-b)/c(x+a)`

Similarly, the side opposite (a,0) has slope `m=(c-0)/(b+a)` , so the slope of this altitude is `m=-(a+b)/c` . Using the point-slope form with this slope through the point (a,0) gives `y=-(a+b)/c(x-a)`

Finally, since the third side lies along the x-axis, the equation of the altitude is `x=b`

Plugging `x=b` into the first equation gives `y=(a^2-b^2)/c` , and plugging `x=b` into the second equation gives `y=-(b^2-a^2)/c=(a^2-b^2)/c`

` `` `Thus the point` ` ` ``(b,(a^2-b^2)/c)` lies on all three lines. Therefore they are concurrent.

I can't make sketch here to show the proof clearly but I'll describe the sketch so that you can follow by making the sketch on your paper while reading this.

Let's draw a triangle with sides **a**,** b**, **c**. Pls. put c on the longest of the three sides. Now name the vertices as **A**, **B**, **C** with **A** opposite side **a**; **B **opposite side **b** and **C** opposite side** c**.

Now draw the first altitude perpendicular to side **c**.. This will pass through vertex **C**. We will call this as "**altC**". Put a small square at the intersection of **altC** and side **c**, to indicate right angle and mark this intersection as point **F**.

Next draw the second altitude perpendicular to side **a**. This line will pass through vertex **A**. We will call this as "**altA**". Put a small square at the intersection of **altA** and side **a**, to indicate right angle and mark this intersection as point **D**. *(Note: If angle C is acute, then point D is on side a in the triangle, else point D is on the extension of side a outside the triangle.)*

Finally draw the last altitude through vertex **B** and perpendicular to side **b** and we will call this as "**altB**". Mark the intersection of **altB** and side **b** as point **E** and put small square on it. *(Again if angle C is acute, then point E is on side b in the triangle, else point E is on the extension of side b outside the triangle.)*

Now locate the intersection of **altA** and **altC** and mark this as point **O1**, then locate also the intersection of **altB** and **altC** and mark as point **O2**. then locate also the intersection of **altA** and **altB** and mark as point **O**. (*For your guide if angle C is acute, then point O1, O2 and O is inside the triangle, else point O1, O2 and O are outside the triangle.)*

*If your drawing is perfect the you'll see that the altitudes are concurrent. But it's better if they're not. so that we a triangle of error O1O2O*

**O1, O2 **and **O** are the same point, if they are concurrent and **FO1=FO2**. So we can make a proposition that:

*The altitudes of a triangle are concurrent because distance FO1 and distance FO2 are equal.*

Proof:

From right triangle AFC

FC = bsinA; AF = bcosA

From right triangle BFC

FC = asinB; BF = acosB

bsinA = asinB

b = asinB/sinA

But <ABD = 90-A, thus <FO2B = A

Also <BAE = 90-B, thus <FO1A = B

From right triangle AFO1

tanFO1A = AF/FO1

FO1 = AF/ tanFO1A

FO1 = bcosA/tanB

FO1 = (asinB/sinA)cosA /tanB

FO1 =acotAcosB

From right triangle BFO2

tan FO2B = BF/FO2

FO2 = BF/tan FO2B

FO2 = acosB/tanA

FO2 = acosBcotA

Distance FO1 is equal to distance FO2, therefore

*The altitudes of a triangle are concurrent*