2 Answers | Add Yours
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.
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
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
We’ve answered 333,869 questions. We can answer yours, too.Ask a question